#1**-1 **

how did you upload a picture it wont worrk for me cant figure it out msg me

Mathatesme Mar 13, 2019

#5**+1 **

\(\left .\sum \limits_{k=1}^n ~k^2 = \dfrac 1 6 n (1+n)(1+2n) \right |_{n=2016} = \\ \dfrac 1 6 (2016)(2017)(4033) \)

\(\dfrac 1 6 (2016)(2017)(4033) \pmod{17} = \\ (336 \pmod {17})(2017 \pmod{17})(4033 \pmod{17}) \pmod{17} = \\ 13 \cdot 11 \cdot 4 \pmod{17} = \\ 572 \pmod{17} = 11\)

.Rom Mar 13, 2019

#8**0 **

Wait, how did you get the first summation? Is there a formula or something?

dgfgrafgdfge111
Mar 14, 2019

#10**+2 **

**2016 NS 28**

\(\begin{array}{|ll|} \hline 0^2+1^2+2^2+\ldots +7^2+8^2 &+9^2 +10^2+\ldots +15^2+16^2 \\ +17^2+18^2+19^2+\ldots +24^2+25^2 &+26^2+27^2\ldots +32^2+33^2 \\ \dots \\ +2006^2+2007^2+2008^2+\ldots +2013^2+2014^2 &+2015^2+2016^2 \pmod{17}\\ =\\ \color{red}\underbrace{0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}\underbrace{+9^2 +10^2+11^2+\ldots +15^2+16^2}_{=0\pmod{17}} \\ \color{red}\underbrace{+0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}\underbrace{+9^2 +10^2+11^2+\ldots +15^2+16^2}_{=0\pmod{17}} \\ \color{red}\dots \\ \color{red}\underbrace{+0^2+1^2+2^2+\ldots+7^2+8^2}_{=0\pmod{17}}&\color{red}+9^2+10^2 \pmod{17} \\\\ =9^2+10^2 \pmod{17}\\ =11 \pmod{17}\\ \hline \end{array}\)

heureka Mar 14, 2019